Answer
(a) unchanged
(b) increase
(c) decrease
(c) increase
Work Step by Step
(a) Since the electric field does not depend on the separation $d$, it remains the same.
(b) We know that $C=\frac{\epsilon_{\circ}A}{d}$. This shows that if $d$ is increased then $C$ will decrease. We also know that $\Delta V=\frac{q}{C}$, thus for the decreased value of $C$, $\Delta V$ will increase because it is inversely proportional to $C$.
(c) Since $C=\frac{\epsilon_{\circ}A}{d}$, this equation shows that separation and capacitance are inversely proportional. Hence, if separation is increased then capacitance will decrease.
(d) We know that energy stored in the capacitor is given as $U=\frac{1}{2}CV^2$.
From part (b), it is clear that $\Delta V$ increases, if the separation is increased and as voltage is directly proportional to the energy $U$, the energy stored will increase.
